The Monty Hall Problem is a probability puzzle named after the host of the television game show “Let’s Make a Deal,” Monty Hall. The problem can be stated as follows:
- You are a contestant on a game show. In front of you, there are three doors. Behind one of the doors is a car (the prize you want), and behind the other two doors are goats.
- You choose one of the doors, let’s say Door #1.
- The host, Monty Hall, who knows what is behind each door, opens one of the other two doors, revealing a goat. For example, let’s say he opens Door #3, and you see a goat.
- Now, Monty gives you a choice: You can stick with your original choice (Door #1), or you can switch to the other unopened door (Door #2).
The question is: What should you do to maximize your chances of winning the car? Should you stick with your initial choice, switch, or does it not matter?
The counterintuitive answer is that you should always switch. Here’s the reasoning behind it:
- When you initially pick a door, there’s a 1/3 chance that the car is behind your chosen door and a 2/3 chance that the car is behind one of the other doors.
- If Monty opens one of the other doors to reveal a goat, the 2/3 chance “moves” to the remaining unopened door. So, if you switch, you are more likely to win the car (2/3 chance), whereas if you stick with your initial choice, your chance remains at 1/3.
This might seem counterintuitive at first, as it goes against our initial instinct to think that once Monty reveals a goat, the probability is evenly distributed between the two remaining doors. However, the key is in recognizing that Monty’s action imparts information about the doors, and the initial probabilities are preserved.
The Monty Hall Problem is a classic example in probability theory and often serves as a fascinating illustration of how our intuition can sometimes lead us astray in understanding probability.